For context, this is from a quantum mechanics lecture in which we were considering continuous eigenvalues of the position operator.
Starting with the position eigenvalue equation,
$$\hat{x}\,\phi(x_m, x)=x_m\phi(x_m,x)$$
where $x_m$ is the eigenvalue and $\phi(x_m, x)$ are the continuous eigenfunctions of the position operator $\hat x$.
The professor wrote that $\phi(x_m, x)=\delta(x-x_m)$ and stated that this is because the eigenbasis is continuous.
But then he wrote the following
$$\begin{align}\int_{-\infty}^{\infty}\phi^*(x_m,x)\phi({x_m}',x)\,dx &=\int_{-\infty}^{\infty}\delta(x_m-x)\delta({x_m}'-x)\,dx \tag{1}\\ &=\color{red}{\delta(x_m-{x_m}')}\end{align}$$
I don't understand how the expression in $(1)$ can be equal to the expression in red. I do understand that $$\begin{align}\int_{-\infty}^{\infty}\delta(x_m-x)\phi({x_m}',x)\,dx &=\phi({x_m}',x_m)\tag{2}\\&=\color{#080}{\delta(x_m-{x_m}')}\end{align}$$
since the integral 'sifts' out the only value of $x$ where the argument of the Dirac delta function is zero (at $x_m$). I have applied this sifting property for one Dirac delta function (as in $(2)$).
But I don't understand how this works when there are two Dirac deltas in the integrand, $(1)$.
By my logic, I think it should sift out each value, one at a time, so I think that $(1)$ should be
$$\int_{-\infty}^{\infty}\delta(x_m-x)\delta({x_m}'-x)\,dx =\delta(x_m)+\delta({x_m}')$$
where the results of the integration are added, since the values $x_m$ and ${x_m}'$ are sifted out one after the other, depending on which of $x_m$ and ${x_m}'$ are larger (here I assumed ${x_m}'\gt x_m$).
Could someone please derive or explain why equation $(1)$ is true, or give me any hints to help me understand it.
Best Answer
Intuitively, $\delta(x_m-x)$ is non-zero only when $x_m=x$. Likewise $\delta(x_m'-x)$ is non-zero only when $x_m'=x$. So the factor $\delta(x_m-x) \, \delta(x_m'-x)$ is non-zero only when $x_m=x=x_m'.$ After integration w.r.t. $x$ we still must have $x_m=x_m'$ which makes the result $\delta(x_m-x_m')$ quite natural.
More formally, recall the formula $\int_{-\infty}^{\infty} f(x) \, \delta(x_0-x) \, dx = f(x_0)$. Apply this to the given integral with $x_0=x_m'$ and $f(x) = \delta(x_m-x).$ Then the result $\delta(x_m-x_m')$ falls out.